I’ve been a bit busy lately, and so I missed last week posting. So what I’ve decided to do is to take the talks I’ve given in various graduate student seminars over the last year or so and convert them into posts. This one is a particularly tough prospect, as the talk didn’t go very well. I’m following a paper of Hamilton’s titled “Ricci Flow on Surfaces” and only present the high genus case. Comments and (especially!) corrections are encouraged.

Let M be a compact surface, g_{ij} a Riemannian metric and R the Ricci curvature of g_{ij}. Then the Ricci Flow equation is given by \frac{\partial}{\partial t}g_{ij}=(r-R)g_{ij} where r is the average value of R. Ricci Flow can be used to prove many classical result, including the Uniformization Theorem, that every Riemann surface admits a metric of constant curvature.

This is actually slightly different from the standard Ricci flow, it is what is called the volume-preserving Ricci flow, as, unlike the standard, it doesn’t cause Spheres to shrink. To show this, set \mu=\sqrt{\det g_{ij}}, and see \frac{\partial}{\partial t}\mu=(r-R)\mu, so if A is total area, \frac{d}{dt}A=\frac{d}{dt}\int 1d\mu=\int(r-R)d\mu=0, because r=\int Rd\mu/\int 1d\mu. We also note that the flow is pointwise a multiple of the metric, and so preserves conformal structure.

In face, we can see that \int Rd\mu=4\pi \chi(M) by the Gauss-Bonnet Theorem, and so r=4\pi \chi(M)/A.

The curvature is a function of the metric, and so it will itself satisfy some evolution equation. This equation happens to be \frac{\partial R}{\partial t}=\Delta R+R^2-rR. (The gist is that for surfaces R=R_{1212}, and we can then take the time derivative, and substitute in the Ricci Flow equation for \dot{g}, but doing it out is messy. Consider it an exercise.)

By the Maximum Principle, we have that if R\geq 0 at the start, it will remain so for all time. Likewise, if R\leq 0, it will remain so. Thus, both positive and negative curvature are preserved for surfaces.

We will only prove that R\geq 0, and the other case follows from a similar argument. We will proceed by contradiction, and assume that at some time, for some point p, R<0. Let t_0 be the first time such that R(t_0,p)=0 and R(t_0+\epsilon,p)<0 for some \epsilon>0. We define m(t)=\min_{p\in M}(R(t)). At t_0, we see that \partial_t R\leq 0 at p, by our assumption. Now, at the same point and time, we look at \Delta R+R^2-rR. R^2-rR=0, as R=0, and so we look at \Delta R. As we are at a minimum with m(t), \Delta R\geq 0, and as \partial_tR=\Delta R+R^2-rR, this means we have a contradiction (citing the maximum principle, which gives one of these a strict inequality).

If R\leq 0, we can strengthen this, and get that if -C\leq R\leq -\epsilon<0 at the start, then it remains so, and re^{-\epsilon t}\leq r-R\leq C e^{rt}, so R approaches r exponentially.

To see it, let R_{\max} be the maximum of R. Then it satisfies \frac{d}{dt}R_{\max}\leq R_{\max}(R_{\max}-r)\leq -\epsilon (R_{\max} -r) and if R_{\min} is the minimum of R, it satisfies \frac{d}{dt}R_{\min}\geq R_{\min}(R_{\min}-r)\geq r(R_{\min}-r)

This implies immediately that on a compact surface, if R<0, a solution exists for all time and converges exponentially to a metric of constant negative curvature. For R\geq 0, things are harder, as R=r is a repulsive fixed point of \frac{dR}{dt}=R^2-rR. The best this method gives is the following:

If r>0 and R/r\geq c>0 at the start, then c<1 and for all time \frac{R}{r}\geq\frac{1}{1-(1-\frac{1}{c})e^{rt}} and if r>0 and R/r\leq C at the start, then C>1 and \frac{R}{r}\leq \frac{1}{1-(1-\frac{1}{c})e^{rt}}, at least for t<\frac{1}{r}\log\frac{c}{c-1}.

These don’t give good bounds, as the lower bound goes to zero at infinity, and the upper bound goes to infinity in finite time.

If R>0 anywhere, we need better methods. We first define the potential f to be the solution to \Delta f=R-r with mean value zero. This equation can always be solved as R-r has mean value zero, and the solution is unique up to a constant, so f can have mean value zero. Then f satisfies the following equation:

\frac{\partial f}{\partial t}=\Delta f+rf-b where b=\int |Df|^2d\mu/\int 1d\mu with b a constant on the surface and merely relying on time.

As \Delta f=R-r, we can get \Delta \frac{\partial f}{\partial t}=\Delta(\Delta f+rf) by differentiating, and so \frac{\partial f}{\partial t}=\Delta f+rf-b for some number b which is only a function of time. b can be computed from the relation \int fd\mu=0.

To make more progress, we will need the function h=\Delta f+|Df|^2 and the tensor M_{ij}=D_iD_j f-\frac{1}{2}\Delta f\cdot g_{ij}, that is, the trace-free part of the second covariant derivative of f.

We get the following equation for h, where |M_{ij}|^2=M^{ij}M_{ij}:

\frac{\partial h}{\partial t}=\Delta h-2|M_{ij}|^2+rh

If h\leq C at the start, then h\leq Ce^{rt} for all time. This is of value, as R=h-|Df|^2+r, so now R\leq Ce^{rt}+r, which gives us a bound on R from above which goes to \infty as t increases if r>0. We can also get a lower bound, if r\geq 0 and the minimum of R is negative, it increases. If r\leq 0 and the minimum of R is less than r, it increases. This gives us that for any initial metric, there is a C such that -C\leq R\leq Ce^{rt}+r. Thus, the Ricci Flow has solutions for all time for any initial metric. In fact, if r\leq 0, then R remains bounded both above and below such that when r<0, R<0 for large time. Applying our earlier result for the situation when R<0, this gives the following result:

On a compact surface with r<0, for any initial metric the solution exists for all time and converges to a metric with constant negative curvature.

And as the Gauss-Bonnet Theorem relates r to the Euler characteristic, we have proved that on a riemann surface with g\geq 2 there exists a metric of constant curvature. The other finitely many cases can be checked by hand (in fact, restricting to compact surfaces, the only things remaining are the sphere, torus, projective plane and klein bottle).